Chapter 8 of 8

Updated 6 August 2026

NCERT Solutions

Development of Surfaces — NCERT Solutions

CBSE · Class 11 · Engineering Graphics

NCERT Solutions for Development of Surfaces, CBSE Class 11 Engineering Graphics: 15 textbook questions solved step by step.

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15 Questions Solved · 2 Sections

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Short Questions

1In drawing the development of objects, true lengths are used. (True/False)Show solution

Answer: True

In the development of surfaces, every line drawn on the development must represent the true length of the corresponding line on the actual surface of the object. If a line is not in its true length (i.e., it is foreshortened in the given views), its true length must first be determined before it can be used in the development. This is a fundamental principle of surface development.

2True length of slant edge need not be known to draw a radial development. (True/False)Show solution

Answer: False

To draw a radial line development (used for pyramids and cones), the true length of the slant edge (or slant generator in the case of a cone) is absolutely essential. The slant edge serves as the radius of the arc used to draw the development. Without knowing the true length of the slant edge, it is impossible to construct an accurate radial development.

3Every line on a development must be equal to the true length of that line on the actual surface. (True/False)Show solution

Answer: True

This is the fundamental rule of surface development. A development is the unfolding of all the surfaces of a solid onto a single flat plane. For the development to be accurate and usable (e.g., for sheet-metal work), every line — whether it is a base edge, lateral edge, or slant edge — must be drawn at its true length. Any foreshortened line must be converted to its true length before being used in the development.

4Name the methods of development of right solids.Show solution

Answer:

The following methods are used for the development of right solids:

Parallel Line Development — Used for prisms and cylinders, where the lateral edges or generators are parallel to each other. The lateral surface is unrolled by drawing parallel lines.

Radial Line Development — Used for pyramids and cones, where the lateral edges or generators meet at a common apex. The slant edge (or slant height) is used as the radius to draw the development.

Triangulation Development — Used for transition pieces and oblique solids whose surfaces are divided into triangles. Each triangle is drawn in its true shape to build up the development.

Approximate Development — Used for double-curved surfaces such as spheres, which cannot be developed exactly. The surface is approximated by dividing it into zones or lunes.

5To develop the surfaces of pyramids, it is necessary to find ________ of the slant edges when they are not parallel to reference plane.Show solution

Answer: True Length

Complete sentence: To develop the surfaces of pyramids, it is necessary to find the true length of the slant edges when they are not parallel to the reference plane.

Explanation: In the orthographic views (front view and top view), slant edges of a pyramid appear foreshortened if they are inclined to the reference plane. Since the development requires every edge to be drawn at its actual (true) length, the true length of each slant edge must be determined — typically by the rotation method or by constructing a right triangle — before the radial line development can be drawn.

Assignments

1Draw the development of a cube of side 50 mm.Show solution

Given: Cube of side = 50 mm

Concept: A cube has 6 equal square faces. Its development (parallel line method) is a cross-shaped flat pattern.

Steps to Draw the Development:

Step 1 – Understand the solid:
A cube has 6 faces: Top, Bottom, Front, Back, Left, Right — all squares of side 50 mm.

Step 2 – Draw the lateral surface (four side faces in a row):

Draw a horizontal baseline.
Mark points $A$ , $B$ , $C$ , $D$ , $E$ along the baseline such that $AB = BC = CD = DE = 50$ mm.
This gives the stretch-out line of length $4 \times 50 = 200$ mm.
Draw vertical lines of height 50 mm at each point.
This produces four squares side by side: representing the Front, Right, Back, and Left faces.

Step 3 – Add the Top face:

Above the second square (Right face or any chosen face), draw one more square of side 50 mm.

Step 4 – Add the Bottom face:

Below the same reference square, draw one more square of side 50 mm.

Result:
The complete development is a cross (plus-sign) shape consisting of 6 squares, each of side 50 mm.

Total surface area $= 6 \times 50^2 = 6 \times 2500 = 15000 \text{ mm}^2$

$\boxed{\text{Development: A cross-shaped pattern of 6 squares, each } 50 \text{ mm} \times 50 \text{ mm}}$

2Draw the development of a Triangular pyramid of base edge 30 mm and height of 60 mm.Show solution

Given:

Base edge of triangular pyramid = 30 mm
Height = 60 mm

Concept: A triangular pyramid (tetrahedron with triangular base) has 1 triangular base and 3 triangular lateral faces. The radial line method is used. The true length of the slant edge is required.

Step 1 – Find the True Length of the Slant Edge:

For a regular triangular pyramid:

Base edge $a = 30$ mm
The centroid of an equilateral triangle is at a distance of $\frac{a}{\sqrt{3}} = \frac{30}{\sqrt{3}} \approx 17.32$ mm from each vertex.

True length of slant edge:
$l = \sqrt{h^2 + r^2} = \sqrt{60^2 + 17.32^2} = \sqrt{3600 + 300} = \sqrt{3900} \approx 62.45 \text{ mm}$

Step 2 – Draw the Development:

Mark apex point $O$ .
With radius = true slant edge length $\approx 62.45$ mm, draw arcs from $O$ .
Mark point $A$ on the arc. With radius = base edge = 30 mm, mark $B$ on the arc from $A$ . Similarly mark $C$ from $B$ , and close back to $A$ .
Join $OA$ , $OB$ , $OC$ , $OA$ — these are the three lateral triangular faces.
Attach the equilateral triangular base (side 30 mm) to any one base edge.

Result:
The development consists of 3 congruent isosceles triangles (lateral faces) arranged around apex $O$ , with slant edge $\approx 62.45$ mm and base 30 mm, plus one equilateral triangular base of side 30 mm.

$\boxed{\text{Slant edge} \approx 62.45 \text{ mm; Development = 3 lateral triangles + 1 base triangle}}$

3Draw the development of a square prism of base side 35 mm and axes of 50 mm long.Show solution

Given:

Base side of square prism = 35 mm
Height (axis length) = 50 mm

Concept: A square prism has 4 rectangular lateral faces and 2 square bases. The parallel line method is used.

Step 1 – Draw the Stretch-Out Line:

The perimeter of the square base $= 4 \times 35 = 140$ mm.
Draw a horizontal line (stretch-out line) of length 140 mm.
Mark points $A$ , $B$ , $C$ , $D$ , $A'$ such that $AB = BC = CD = DA' = 35$ mm.

Step 2 – Draw the Lateral Surface:

At each marked point, draw vertical lines of height = 50 mm (the axis length).
Connect the tops of these verticals with a horizontal line.
This gives 4 rectangles, each $35 \text{ mm} \times 50 \text{ mm}$ , representing the four lateral faces.

Step 3 – Attach the Bases:

Attach a square of side 35 mm to the bottom of any one rectangle (for the bottom base).
Attach a square of side 35 mm to the top of any one rectangle (for the top base).

Result:
The complete development is a rectangle of $140 \text{ mm} \times 50 \text{ mm}$ (lateral surface) with two $35 \text{ mm} \times 35 \text{ mm}$ squares attached for the top and bottom.

$\boxed{\text{Stretch-out length} = 140 \text{ mm},\ \text{Height} = 50 \text{ mm}}$

4A hexagonal prism of base side 30 mm and height of 60 mm is resting on its base with its axis perpendicular to the H.P. Develop its surface.

Given:

Base side of hexagonal prism = 30 mm
Height = 60 mm
Axis perpendicular to H.P. (resting on base)

Concept: A hexagonal prism has 6 rectangular lateral faces and 2 hexagonal bases. Parallel line method is used.

Step 1 – Calculate the Stretch-Out Length:
$\text{Perimeter} = 6 \times 30 = 180 \text{ mm}$

Step 2 – Draw the Lateral Development:

Draw a horizontal baseline of length 180 mm.
Mark 7 points: $A, B, C, D, E, F, A'$ at equal intervals of 30 mm.
At each point, erect a vertical line of height 60 mm.
Connect the tops with a horizontal line.
This produces 6 rectangles, each $30 \text{ mm} \times 60 \text{ mm}$ .

Step 3 – Attach the Hexagonal Bases:

Draw a regular hexagon of side 30 mm attached to the bottom edge of the first rectangle (bottom base).
Draw a regular hexagon of side 30 mm attached to the top edge of any rectangle (top base).

Result:
The lateral development is a rectangle of $180 \text{ mm} \times 60 \text{ mm}$ with two regular hexagons (side 30 mm) for the top and bottom bases.

$\boxed{\text{Stretch-out} = 180 \text{ mm},\ \text{Height} = 60 \text{ mm},\ \text{Bases: Regular hexagons of side 30 mm}}$

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5A cylinder having diameter of 40 mm and 65 mm high is kept on its base. Develop its surface.

Given:

Diameter of cylinder $d = 40$ mm, so radius $r = 20$ mm
Height $h = 65$ mm

Concept: The lateral surface of a cylinder, when unrolled, forms a rectangle. The width equals the circumference of the base circle and the height equals the height of the cylinder. Parallel line method is used.

Step 1 – Calculate the Circumference (Stretch-Out Length):
$L = \pi d = \pi \times 40 = 125.66 \text{ mm} \approx 125.7 \text{ mm}$

Step 2 – Draw the Lateral Development:

Draw a rectangle of length $125.7$ mm and height $65$ mm.
This rectangle represents the complete lateral (curved) surface of the cylinder.

Step 3 – Attach the Circular Bases:

Draw a circle of diameter 40 mm ( $r = 20$ mm) attached to the bottom edge of the rectangle (bottom base).
Draw a circle of diameter 40 mm attached to the top edge of the rectangle (top base).

Result:
The development consists of:

One rectangle: $125.7 \text{ mm} \times 65 \text{ mm}$ (lateral surface)
Two circles of diameter 40 mm (top and bottom bases)

$\boxed{\text{Lateral surface: Rectangle } \pi \times 40 \approx 125.7 \text{ mm} \times 65 \text{ mm}}$

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6Draw the development of a pentagonal prism of base side 30 mm and height of 55 mm.

Given:

Base side of pentagonal prism = 30 mm
Height = 55 mm

Concept: A pentagonal prism has 5 rectangular lateral faces and 2 pentagonal bases. Parallel line method is used.

Step 1 – Calculate the Stretch-Out Length:
$\text{Perimeter} = 5 \times 30 = 150 \text{ mm}$

Step 2 – Draw the Lateral Development:

Draw a horizontal baseline of length 150 mm.
Mark 6 points: $A, B, C, D, E, A'$ at equal intervals of 30 mm.
At each point, erect a vertical line of height 55 mm.
Connect the tops with a horizontal line.
This produces 5 rectangles, each $30 \text{ mm} \times 55 \text{ mm}$ .

Step 3 – Attach the Pentagonal Bases:

Draw a regular pentagon of side 30 mm attached to the bottom edge (bottom base).
Draw a regular pentagon of side 30 mm attached to the top edge (top base).

Note on drawing a regular pentagon (side 30 mm):
Interior angle of regular pentagon $= 108°$ . Draw using compass and protractor or geometric construction.

Result:
The lateral development is a rectangle of $150 \text{ mm} \times 55 \text{ mm}$ with two regular pentagons (side 30 mm) for the bases.

$\boxed{\text{Stretch-out} = 150 \text{ mm},\ \text{Height} = 55 \text{ mm}}$

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7Draw the development of a triangular pyramid of base side 35 mm and axis of 60 mm.

Given:

Base side of triangular pyramid = 35 mm
Axis (height) = 60 mm

Concept: A regular triangular pyramid has 1 equilateral triangular base and 3 congruent isosceles triangular lateral faces. Radial line method is used.

Step 1 – Find the True Length of the Slant Edge:

For a regular triangular pyramid, the distance from the centroid to a vertex (circumradius of base):
$R = \frac{a}{\sqrt{3}} = \frac{35}{\sqrt{3}} \approx 20.21 \text{ mm}$

True length of slant edge:
$l = \sqrt{h^2 + R^2} = \sqrt{60^2 + 20.21^2} = \sqrt{3600 + 408.4} = \sqrt{4008.4} \approx 63.31 \text{ mm}$

Step 2 – Draw the Development:

Mark apex $O$ .
With radius $= l \approx 63.31$ mm, draw an arc.
Mark point $A$ on the arc. With compass set to base edge = 35 mm, step off points $B$ , $C$ , and back to $A$ along the arc.
Join $OA$ , $OB$ , $OC$ , $OA$ — forming 3 lateral triangular faces.
Attach the equilateral triangular base (side 35 mm) to one of the base edges.

Result:
Development = 3 isosceles triangles (slant edge $\approx 63.31$ mm, base 35 mm) + 1 equilateral triangle (side 35 mm).

$\boxed{\text{Slant edge} \approx 63.31 \text{ mm};\ \text{Base edge} = 35 \text{ mm}}$

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8A pentagonal pyramid of base side 25 mm and height of 50 mm is kept on its base. Develop its surface.

Given:

Base side of pentagonal pyramid = 25 mm
Height = 50 mm

Concept: A regular pentagonal pyramid has 1 pentagonal base and 5 congruent isosceles triangular lateral faces. Radial line method is used.

Step 1 – Find the True Length of the Slant Edge:

Circumradius of regular pentagon with side $a = 25$ mm:
$R = \frac{a}{2\sin(36°)} = \frac{25}{2 \times 0.5878} = \frac{25}{1.1756} \approx 21.27 \text{ mm}$

True length of slant edge:
$l = \sqrt{h^2 + R^2} = \sqrt{50^2 + 21.27^2} = \sqrt{2500 + 452.4} = \sqrt{2952.4} \approx 54.34 \text{ mm}$

Step 2 – Draw the Development:

Mark apex $O$ .
With radius $= l \approx 54.34$ mm, draw an arc.
Mark point $A$ on the arc. With compass set to 25 mm, step off $B$ , $C$ , $D$ , $E$ , and back to $A$ — giving 5 equal divisions.
Join $O$ to each point: $OA$ , $OB$ , $OC$ , $OD$ , $OE$ , $OA$ — forming 5 lateral triangular faces.
Attach the regular pentagonal base (side 25 mm) to one base edge.

Result:
Development = 5 isosceles triangles (slant edge $\approx 54.34$ mm, base 25 mm) + 1 regular pentagon (side 25 mm).

$\boxed{\text{Slant edge} \approx 54.34 \text{ mm};\ \text{Base edge} = 25 \text{ mm}}$

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9Draw the development of a pentagonal pyramid of base edge 30 mm and 60 mm height.

Given:

Base edge of pentagonal pyramid = 30 mm
Height = 60 mm

Concept: A regular pentagonal pyramid has 1 pentagonal base and 5 congruent isosceles triangular lateral faces. Radial line method is used.

Step 1 – Find the True Length of the Slant Edge:

Circumradius of regular pentagon with side $a = 30$ mm:
$R = \frac{a}{2\sin(36°)} = \frac{30}{2 \times 0.5878} = \frac{30}{1.1756} \approx 25.52 \text{ mm}$

True length of slant edge:
$l = \sqrt{h^2 + R^2} = \sqrt{60^2 + 25.52^2} = \sqrt{3600 + 651.3} = \sqrt{4251.3} \approx 65.20 \text{ mm}$

Step 2 – Draw the Development:

Mark apex $O$ .
With radius $= l \approx 65.20$ mm, draw an arc.
Mark point $A$ on the arc. With compass set to base edge = 30 mm, step off $B$ , $C$ , $D$ , $E$ , and back to $A$ along the arc — 5 equal chords.
Join $O$ to each base point: $OA$ , $OB$ , $OC$ , $OD$ , $OE$ , $OA$ — forming 5 lateral triangular faces.
Attach the regular pentagonal base (side 30 mm) to one of the base edges.

Result:
Development = 5 isosceles triangles (slant edge $\approx 65.20$ mm, base 30 mm) + 1 regular pentagon (side 30 mm).

$\boxed{\text{Slant edge} \approx 65.20 \text{ mm};\ \text{Base edge} = 30 \text{ mm}}$

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10Develop the surface of a cone of base diameter 50 mm and 60 mm axis.

Given:

Base diameter $d = 50$ mm, so radius $r = 25$ mm
Axis (height) $h = 60$ mm

Concept: The lateral surface of a cone, when unrolled, forms a sector of a circle. The radius of the sector equals the slant height $l$ of the cone, and the arc length of the sector equals the circumference of the base circle. Radial line method is used.

Step 1 – Find the Slant Height (True Length of Generator):
$l = \sqrt{r^2 + h^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65 \text{ mm}$

Step 2 – Find the Sector Angle $\theta$ :

The arc length of the sector = circumference of base circle:
$\text{Arc length} = 2\pi r = 2\pi \times 25 = 50\pi \text{ mm}$

The sector angle:
$\theta = \frac{\text{Arc length}}{l} \times \frac{180°}{\pi} = \frac{50\pi}{65} \times \frac{180°}{\pi} = \frac{50 \times 180°}{65} = \frac{9000°}{65} \approx 138.46°$

Alternatively: $\theta = \frac{r}{l} \times 360° = \frac{25}{65} \times 360° \approx 138.46°$

Step 3 – Draw the Development:

Mark apex $O$ .
Draw a line $OA$ of length $l = 65$ mm.
With $O$ as centre and radius 65 mm, draw an arc subtending angle $\theta \approx 138.46°$ at $O$ .
Mark the end point $A'$ of the arc.
Join $OA'$ .
The sector $OAA'$ represents the complete lateral surface of the cone.
Attach a circle of radius 25 mm (diameter 50 mm) to the base edge for the base.

Result:

Slant height $l = 65$ mm
Sector angle $\theta \approx 138.46°$
Development = sector of radius 65 mm and angle $138.46°$ + circle of radius 25 mm

$\boxed{l = 65 \text{ mm},\ \theta = \frac{r}{l} \times 360° = \frac{25}{65} \times 360° \approx 138.46°}$

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